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morpeh [17]
3 years ago
13

Marie mixes 20 liters of 12% acid solution with a 36% acid solution to make a 20% acid solution. How many liters of 36% solution

did she use?
Mathematics
1 answer:
svet-max [94.6K]3 years ago
6 0
She used 10 liters of the 36% solution.
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Steve sells half of his comic books and then buys 6 more now he has 14 comic books
ddd [48]

Let b be the initial amount of books. Steve sells half of his books, so the number of Steve's books become

b \to \dfrac{b}{2}

Then, he buys 6 more, so the number increases by 6:

\dfrac{b}{2} \to \dfrac{b}{2}+6

This number is now 14, so you have

\dfrac{b}{2}+6 = 14

Subtract 6 from both sides:

\dfrac{b}{2} = 8

Multiply both sides by 2:

b = 16

4 0
3 years ago
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4/5 of a number is 32​
andrew11 [14]

Answer:

x = 40

Step-by-step explanation:

Step 1: Write out an equation

4/5(x) = 32

Step 2: Solve by dividing both sides by 4/5

x = 32/(4/5)

<em>Use KCF (Keep Change Flip) if you do not have a calc.</em>

x = 40

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Anya recorded the height in inches of 10 students in her class. The results are shown below.
Juliette [100K]
D, add all your answers than divide by how many numbers there are
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Is FGH ~ JKL? If so, identify the similarity postulate or theorem that applies.
Leokris [45]
Yes, for triangles to similar two angles have to be the same, in this case, they both have 55 and 30 degrees), this makes the third angle the same. Triangles angles add up to 180 degrees and if you subtract 30 and 55 from 180..
180-30=150-55=95
95+30+55=180

8 1
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Pleaseee helppp with thissss asapppp
Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

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